How to prove a limit with a recurrence? $s_1 = 1$ and $s_{n+1} = \dfrac{s_n + 1}{3}$ for $n \in \Bbb N$.
How do you find $\displaystyle \lim_{x\to \infty} s_n$?
Then how do you prove that the value is the limit using the definition of the limit of a sequence?  $|s_n - s| < \epsilon$ for $n$ sufficiently large. 
I know the limit is $\frac{1}{2}$. I'm just having trouble with the recurrence aspect.
 A: First, we need that this sequence actually converges. It's not too hard to show that $\{s_n\}$ is a decreasing sequence, which is clearly bounded below by 0. Thus, $\{s_n\}$ converges, because bounded, monotonic sequences converge.
Now, we use the fact that $$ \lim_{n\to \infty} s_n = \lim_{n\to \infty} s_{n+1}.$$ If $L = \lim_{n\to \infty} s_n$, then
$$
 L = \frac{L+1}{3}
$$
implies $L=\frac{1}{2}$.
A: Below are the steps:


*

*Use induction to show that $s_n > \dfrac12$.

*Now again use induction to show that $s_n$ is a monotonically decreasing sequence, i.e., $s_{n+1} < s_n$.

*Now use the completeness of $\mathbb{R}$ to conclude that a monotonically decreasing sequence bounded below converges to a real number.

*Now let $\lim_{n \to \infty} s_n = L$. Use the properties of limit, to conclude that $\lim_{n \to \infty} s_{n+1} = L$ and also $\lim_{n \to \infty} \dfrac{s_n+1}3 = \dfrac{L+1}3$.

*Now conclude that $$L = \dfrac{L+1}3 \implies L = \dfrac12$$

A: As you know the limit is $\frac 12$, define $t_n=s_n-\frac 12$.  Plug that in to your recurrence.  It should be easy to get a formula for $t_n$ as a function of $t_0$, which you can use to find a formula for $s_n-\frac 12$ based on $s_1-\frac 12$ that you can use.
A: I know of two ways to find this limit.
1) I would first prove that it's convergent (prove it's cauchy or prove it's increasing and bounded above or decreasing and bounded below). Next I would then pass the limit through the recursion relation, I'll explain what I mean by this next. We know, since the sequence is convergent that
$$
\exists \; L \in \mathbb{R} : \lim_{n \to \infty} s_n = L
$$
Next we observe that if $s_n \to L$ then $s_{n+1} \to L$ (considering that the limit is taken as $n \to \infty$ the $+1$ doesn't really make a difference, it still will approach the same $L$). So we can then see
$$
s_{n+1} = \frac{s_n + 1}{3} \implies L = \frac{L + 1}{3} \iff 2L = 1 \iff L = \frac{1}{2} 
$$
2) The next (and more difficult in my opinion) way is to prove it by the definition by rewriting $s_n$ as a non-recursive sequence only dependent on $s_1$ (similar to what Ross says)
A: From $s_{n+1} = (s_n + 1)/3$,
we want to eliminate the "$+1$"
to get a simpler recurrence.
Let $s_n = t_n+c$.
Then
$t_{n+1}+c = (t_n+c+1)/3
= t_n/3 + (c+1)/3
$
so $t_{n+1} = t_n/3 -c + (c+1)/3
=t_n/3 + (1-2c)/3
$.
If we choose $c = 1/2$,
then $t_{n+1} = t_n/3$,
so that $t_n = t_0/3^n \to 0$ as $n \to \infty$.
Thus $s_n = t_n + 1/2 \to 1/2$.
A: Solve the recurrence, the limit falls out of the solution. By "generatingfuctionology", define $S(z) = \sum_{n \ge 0} s_n z^n$, multiply your recurrence by $z^n$ and add for $n \ge 0$ to get:
$$
\frac{S(z) - s_0}{z} = \frac{1}{3} S(z) + \frac{1}{3} \cdot \frac{1}{1 - z}
$$
From the recurrence "backwards" you have $s_0 = 2$, so:
$$
S(z) = \frac{1}{2} \cdot \frac{1}{1 - z} + \frac{3}{2} \cdot \frac{1}{1 - z / 3}
$$
This is two geometric series:
$$
s_n = \frac{1}{2} + \frac{3}{2} \cdot 3^{-n}
$$
Thus:
$$
\lim_{n \to \infty} s_n = \frac{1}{2}
$$
